A jogger travels a route that has two parts. The first is a displacement of 2.90 km due south, and the second involves a displacement that points due east. The resultant displacement + has a magnitude of 4.25 km. (a) What is the magnitude of , and what is the direction of + as a positive angle relative to due south? (b) Suppose that - had a magnitude of 4.25 km. What then would be the magnitude of , and what is the direction of - relative to due south?
X = ?
Y = -2.9 km.
R = 4.25 km = Resultant.
X^2 + Y^2 = R^2
X^2 + (-2.9)^2 = (4.25)^2
X^2 = (4.25)^2 - (-2.9)^2 = 9.653
X = 3.1 km
tanA = -2.9/3.1 = -0.93548
A = -43.1 o
A = -43.1 + 360 = 316.9o CCW.
A = 316.9 - 270 = 46.9o East of South.
To solve this problem, we can use the concept of vector addition.
(a) To find the magnitude of the resultant displacement (+), we can use the Pythagorean theorem. The Pythagorean theorem states that for a right triangle, the square of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b). In this case, the two sides are the displacements from the jogger's route.
Let's represent the displacement due south as (-2.90 km) since it is in the opposite direction. Let the displacement due east be x. So, our equation becomes:
(-2.90 km)^2 + x^2 = (4.25 km)^2
Simplifying this equation, we get:
8.41 km^2 + x^2 = 18.06 km^2
Subtracting 8.41 km^2 from both sides, we have:
x^2 = 18.06 km^2 - 8.41 km^2
x^2 = 9.65 km^2
Taking the square root of both sides, we get:
x = 3.11 km (rounded to two decimal places)
Therefore, the magnitude of the displacement due east is 3.11 km.
Now, to find the magnitude of the resultant displacement (+), we use the Pythagorean theorem again:
(+)^2 = (-2.90 km)^2 + (3.11 km)^2
(+)^2 = 8.41 km^2 + 9.65 km^2
(+)^2 = 18.06 km^2
Taking the square root of both sides, we have:
+ = 4.25 km
So, the magnitude of the resultant displacement (+) is 4.25 km.
To find the direction of the resultant displacement (+) as a positive angle relative to due south, we can use trigonometry. Since the displacement due south is in the negative direction, we need to find the angle between the resultant displacement (+) and the negative y-axis.
Let's call this angle θ. We can find θ using the inverse tangent function:
θ = arctan(3.11 km / 2.90 km) ≈ 48.05 degrees
Therefore, the direction of the resultant displacement (+) relative to due south is approximately 48.05 degrees.
(b) If the magnitude of the displacement (-) is also 4.25 km, the process to determine the magnitude of the resultant displacement (-) and its direction relative to due south would be the same as in part (a) above.